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# Summation of finite series Watch

1. If $f(r)=\frac{1}{r(r+1)}$ , obtain and simplify f(r+1)-f(r).

Hence, find $\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2) }$.
2. What have you tried?
3. (Original post by cucumberpj)
...
Are you studying this material as an independent candidate - you seem to be struggling with a wide range of questions so I am wondering if you have actually been taught any of this
4. $f(r)= \frac{1}{r(r+1)}$

$f(r)= \frac{1}{(r+1)(r+2)}$

f(r+1)-f(r)
$= \frac{1}{(r+1)(r+2)}-\frac{1}{r(r+1)}$

$= \frac{-2}{r(r+1)(r+2)}$

then how to find the summation?
5. (Original post by TenOfThem)
Are you studying this material as an independent candidate - you seem to be struggling with a wide range of questions so I am wondering if you have actually been taught any of this
I learnt these all under sequences and series.
These questions seemed tricky for me.
6. (Original post by cucumberpj)
I learnt these all under sequences and series.
These questions seemed tricky for me.
They are
7. (Original post by TenOfThem)
They are
So I hope I can seek help from you all.
I am doing these in limited sources.
8. (Original post by cucumberpj)
So I hope I can seek help from you all.
I am doing these in limited sources.
Try doing it yourself for a few small values of n to see what is happening. There should be a lot of cancelling so if you are doing big calculations you've gone wrong.
9. (Original post by cucumberpj)
$f(r)= \frac{1}{r(r+1)}$

$f(r)= \frac{1}{(r+1)(r+2)}$

f(r+1)-f(r)
$= \frac{1}{(r+1)(r+2)}-\frac{1}{r(r+1)}$

$= \frac{-2}{r(r+1)(r+2)}$

then how to find the summation?
Once you've done that bit, you can immediately write down:

The bit on the right is -2 x the sum you're asked to find, and the bit on the left simplifies enormously - try writing it out for a few values of n if you can't see what's going on!

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